We need to find the size of angle $BAC$ in the given triangle $ABC$. We are given that the angle at vertex $B$ is $40^\circ$ and the external angle at vertex $C$ is $120^\circ$.

1. Problem Description

We need to find the size of angle $BAC$ in the given triangle $ABC$. We are given that the angle at vertex $B$ is $40^\circ$ and the external angle at vertex $C$ is $120^\circ$.

2. Solution Steps

First, we need to find the internal angle at vertex $C$. Since the external angle at $C$ is $120^\circ$, and the internal and external angles at a vertex add up to $180^\circ$, we have:

$angle ACB = 180^\circ - 120^\circ = 60^\circ$

The sum of the internal angles in a triangle is $180^\circ$. Therefore:

$angle BAC + angle ABC + angle ACB = 180^\circ$

$angle BAC + 40^\circ + 60^\circ = 180^\circ$

$angle BAC + 100^\circ = 180^\circ$

$angle BAC = 180^\circ - 100^\circ$

$angle BAC = 80^\circ$

3. Final Answer

The size of angle $BAC$ is $80^\circ$.